The equation of hyperbola $H$ is $\dfrac {(x-4)^{2}}{64}-\dfrac {(y+1)^{2}}{49} = 1$. What are the asymptotes?
Answer: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y+1)^{2}}{49} = - 1 + \dfrac {(x-4)^{2}}{64}$ Multiply both sides of the equation by $49$ $(y+1)^{2} = { - 49 + \dfrac{ (x-4)^{2} \cdot 49 }{64}}$ Take the square root of both sides. $\sqrt{(y+1)^{2}} = \pm \sqrt { - 49 + \dfrac{ (x-4)^{2} \cdot 49 }{64}}$ $ y + 1 = \pm \sqrt { - 49 + \dfrac{ (x-4)^{2} \cdot 49 }{64}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y + 1 \approx \pm \sqrt {\dfrac{ (x-4)^{2} \cdot 49 }{64}}$ $y + 1 \approx \pm \left(\dfrac{7 \cdot (x - 4)}{8}\right)$ Subtract $1$ from both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{7}{8}(x - 4) -1$